Question: The area of a square is increasing at a rate of $20$ square meters per hour. At a certain instant, the area is $49$ square meters. What is the rate of change of the perimeter of the square at that instant (in meters per hour)? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{40}{7}$ (Choice B) B $28$ (Choice C) C $7$ (Choice D) D $2\sqrt{5}$
Solution: Setting up the math Let... $s(t)$ denote the square's side at time $t$, $A(t)$ denote the square's area at time $t$, and $P(t)$ denote the square's perimeter at time $t$. We are given that $A'(t)=20$, We are also given that $A(t_0)=49$ for a specific time $t_0$. We want to find $P'(t_0)$. Relating the measures $P(t)$ and $s(t)$ relate to each other through the formula for the perimeter of a square: $P(t)=4s(t)$ We can differentiate both sides to find an expression for $P'(t)$ : $P'(t)=4s'(t)$ $A(t)$ and $s(t)$ relate to each other through the formula for the area of a square: $A(t)=[s(t)]^2$ We can differentiate both sides to find an expression for $A'(t)$ : $A'(t)=2s(t)s'(t)$ Using the information to solve Let's plug ${A(t_0)}={49}$ into the expression for $A(t_0)$ : $\begin{aligned} {A(t_0)}&=[s(t_0)]^2 \\\\ {49}&=[s(t_0)]^2 \\\\ {7}&={s(t_0)} \end{aligned}$ Let's plug ${A'(t_0)}={20}$ and ${s(t_0)}={7}$ into the expression for $A'(t_0)$ : $\begin{aligned} {A'(t_0)}&=2{s(t_0)}s'(t_0) \\\\ {20}&=2({7})s'(t_0) \\\\ C{\dfrac{10}{7}}&=C{s'(t_0)} \end{aligned}$ Now let's plug $C{s'(t_0)}=C{\dfrac{10}{7}}$ into the expression for $P'(t_0)$ : $\begin{aligned} P'(t_0)&=4C{s'(t_0)} \\\\ &=4\left(C{\dfrac{10}{7}}\right) \\\\ &=\dfrac{40}{7} \end{aligned}$ In conclusion, the rate of change of the perimeter of the square at that instant is $\dfrac{40}{7}$ meters per hour. Since the rate of change is positive, we know that the perimeter is increasing.